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| Chapter | 9. Ray Optics and Optical Instruments |
| Subject | Physics |
| Textbook | Physics II |
| Class | Twelve |
| Category | NCERT Solutions for Class 12 |
Class 12 Physics NCERT Solutions are the excellent practice set for students to ace Physics with better understandings and good marks. With the NCERT Solution for Class 12 Physics, students can confidently answer their exam questions and enjoy the best possible marks in the subject. These solutions are prepared keeping in view the latest syllabus and curriculum. In addition, the solutions are designed keeping in mind the requirements of students. Therefore, students can use these solutions for revision or to score high marks in the final examinations. Let’s get started with the solutions of Chapter 9: Ray Optics and Optical Instruments.
NCERT Solutions for Class 12 Physics Chapter 9
Ray Optics and Optical Instruments Solutions
Exercise
Q 9.1) A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm.
At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Answer)
Height of the candle, h =2.5 cm
Image size=h’
Object distance, u= -27 cm
The radius of the concave mirror, R= -36 cm
Focal length of the concave mirror, 
Image distance=v
∴ v = -54cm
Therefore, to obtain a sharp image, the distance between the screen and the mirror should be 54cm.
Image magnification is:
5cm is the height of the image of the candle. As there is a negative sign, the image is inverted and virtual.
If the candle is moved closer to the mirror, then the screen needs to be moved away from the mirror so as to obtain the image.
Q 9.2) A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm.
Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Answer)
Size of the needle, h1=4.5 cm
Object distance, u=-12 cm
Focal length of the convex mirror, f= 15 cm
Image distance= v
The distance between the needle image and the mirror is 6.7cm and the image is obtained on the other side of the mirror.
Using magnification formula:
2.5cm is the height of the image. As it has a positive sign, the image is erect, virtual and diminished.
As the needle is moved farther from the mirror, the image also moves away from the mirror resulting in a reduction in the size of the image.
Q 9.3) A tank is filled with water to a height of 12.5 cm.
The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Answer)
Actual depth of the needle in water, h1=12.5cm
Apparent depth of the needle in water, h2 =9.4 cm
Refractive Index of water = μ
The value of μ can be obtained as follows:
Hence, 1.33 is the refractive index of water.
Now water is replaced by a liquid of refractive index 1.63
The actual depth of the needle remains the same, but its apparent depth changes.
Let x be the new apparent depth of the needle.
Therefore y= 
The new apparent depth of the needle is 7.67cm. It is observed that the value is less than h2, therefore the needle needs to be moved up to the focus again.
Distance to be moved to focus=9.4-7.67=1.73cm
Q 9.4) Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively.
Predict the angle of refraction in glass when the angle of incidence in water is 45º with the normal to a water-glass interface [Fig. 9.34(c)].
Answer)
For the glass-air interface:
Angle of incidence, i =60°
Angle of refraction, r=35°
From Snell’s law we know that the refractive index of the glass with respect to air is given as:
For the air-water interface:
Angle of incidence, i = 60°
Angle of refraction, r = 47°
From Snell’s law we know that the refractive index of the water with respect to air is given as:
With the help of equations (i) and (ii) the relative refractive index of glass with respect to water can be found:
Angle of incidence, i= 45°
Angle of refraction= r
r can be calculated from Snells’s law as
Therefore, 38.68° is the angle of refraction at the water-glass interface.
Q 9.5) A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm.
What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Answer)
Actual depth of the bulb in water, d1=80 cm =0.8 m
Refractive Index of water, μ = 1.33
I = Angle of incidence
r =Angle of refraction – 90°
As the bulb is used as a point source, the emergent light is considered to be a circle.
Using Snell’s law, we can write the relation for the refractive index of water as:
The relation:
Q 9.6) A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°.
What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Answer)
Angle of minimum deviation, δm = 40
Angle of the prism, A = 60°
Refractive index of water, µ = 1.33
Refractive Index of the material = μ‘
The angle of deviation is related to refractive index (μ‘)
1.532 is the refractive index of the material of the prism
δm is the new angle of minimum deviation since the prism is immersed in water.
The relation shows the refractive index of glass with respect to the water:
10.32° is the new minimum angle of deviation.
Q 9.7) Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature.
What is the radius of curvature required if the focal length is to be 20 cm?
Answer)
Refractive Index of glass, μ=1.55
Focal length of the double-convex lens, f=20 cm
Radius of curvature of one face of the lens=R1
Radius of curvature of the other face of the lens=R2
Radius of curvature of the double-convex lens=R
Therefore, R1 = R and R2 = –R
The value of R is calculated as:
22cm is the radius of curvature of the double-convex lens.
Q 9.8) A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P.
At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
Answer)
The object is virtual and the image formed is real.
Object distance, u= +12 cm
(i) The focal length of the convex lens, f =20 cm
Image distance = v
The image will be formed 7.5cm away from the lens, to the right.
(ii) Focal length of the concave lens, f =-16 cm
Image distance =v
∴ v = 48cm.
Therefore, the image obtained is 48cm away from the lens towards the right.
Q 9.9) An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm.
Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Answer)
Size of the object, h1 = 3cm
Object distance, u = -14 cm
Focal length of the concave lens, f = -21 cm
Image distance = v
Therefore, the image obtained is 8.4cm away and is obtained on the other side of the lens. Since there is negative sign, it is understood that the image is virtual and erect.
The height of the image is 1.8cm.
If the object is moved further away from the lens, then the virtual image will move towards the mirror. As the object distance increases, the image size decreases.
Q 9.10) What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm?
Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Answer)
Focal length of the convex lens, f1=30cm
Focal length of the concave lens, f2= -20cm
Focal length of the system of lenses= f
The equivalent focal length of a system of two lenses in contact is given as:
Therefore f=-60cm
Therefore, 60cm is the focal length of the combination of lenses. Also, the negative sign is an indication that the system of lenses used act as a diverging lens.
Q 9.11) A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15 cm.
How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Answer)
Focal length of the objective lens, f1=2.0 cm
Focal length of the eyepiece, f2=6.25 cm
Distance between the objective lens and the eyepiece, d =15 cm
Answer a)
Least distance of distinct vision, d’ = 25 cm
Image distance for the eyepiece, v2 = -25cm
Object distance for the eyepiece= u2
According to the lens formula, we have the relation:
Image distance for the objective lens, v1 = d + u2 = 15 – 5 = 10cm
Object distance for the objective lens = u1
Magnifying power
Hence, 20 is the magnifying power of the microscope.
Answer b)
The final Image is formed at infinity.
Therefore image distance for the eyepiece, v2 = ∞
Object distance for the eyepiece= u2
According to the lens formula, we have the relation:
Image distance for the objective lens, 
Object distance for the objective lens=u1
Following relation is obtained from the lens formula:
Magnitude of the object distance, ∣ u1 ∣ =2.59 cm
Following is the relation explaining the magnifying power of a compound microscope:
Hence, 13.51 is the magnifying power of the microscope.
Q 9.12) A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0 mm from the objective in sharp focus.
What is the separation between the two lenses? Calculate the magnifying power of the microscope,
Answer)
Focal length of the objective lens, fo = 8 mm = 0.8 cm
Focal length of the eyepiece, fo = 2.5 cm
Object distance for the objective lens, u0 = -9.0 mm = -0.9 cm
Least distance of distant vision, d=25 cm
Image distance for the eyepiece, ve = -d = -25 cm
Object distance for the eyepiece = ue
Using the lens formula, we can obtain the value of ue as:
With the help of lens formula, the value of the image distance for the objective (v) lens is obtained:
vo = 7.2 cm
The separation between two lenses are determined as follows:
= v0 + |ue|
= 7.2 + 2.27
= 9.47 cm
The magnifying power of the microscope is calculated as follows:
Magnifying power, M = Mo × Me
Q 9.13) A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm.
What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Answer)
Focal length of the objective lens, fo = 144 cm
Focal length of the eyepiece, fe = 6.0 cm
The magnifying power of the telescope is given as:
The separation between the objective lens and the eyepiece is calculated as:
Therefore, 24 is the magnifying power of the telescope and the separation between the objective lens and the eyepiece is 150cm.
Q 9.14)
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106 m, and the radius of lunar orbit is 3.8 × 108 m.
Answer)
Focal length of the objective lens, fo = 15 m = 15 × 102 cm
Focal length of the eyepiece, fe = 1.0 cm
Answer a)
The angular magnification of a telescope is given as:
Hence, the angular magnification of the given refracting telescope is 1500.
Answer b)
Diameter of the moon, d = 3.48 × 106 m
Radius of the lunar orbit, r0 = 3.8 × 108 m
Let d’ be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
Therefore 13.74cm is the diameter of the image of the moon formed by the objective lens.
Q 9.15: Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
Answer)
The focal length is negative for a concave mirror.
Therefore f < 0
The object distance (u) is negative when the object is placed on the left side of the mirror.
Therefore u < 0
Lens formula for image distance v is written as:
For a concave mirror, f is negative, i.e., f < 0
For a real object, i.e., which is on the left side of the mirror.
For u between f and 2f implies that 1/u lies between 1/f and 1/2f
1/v is negative
which implies v is negative and greater than 2f. Hence, image lies beyond 2f and it is real.
Answer b)
The focal length is positive for a convex mirror.
Therefore f > 0
The object distance (u) will be negative if the object is placed on the left side of the mirror.
Therefore u < 0
Using mirror formula, we can calculate the image distance v:
Using eqn (ii), we can conclude that:
Thus, the image is obtained on the back side of the mirror, Therefore, it can be concluded that a convex mirror always produces a virtual image irrespective of the object distance.
Answer c)
The focal length is positive for a convex mirror.
Therefore f > 0
The object distance (u) is negative when the object is placed on the left side of the mirror
Therefore u < 0
For image distance v, we have the mirror formula:
But we have u<0
Therefore, the image is formed between the focus and the pole and is diminished.
Answer d)
The focal length of the concave mirror is negative.
Therefore f < 0
The object distance, u is negative for an object that is placed on the left side of the mirror.
Therefore u < 0
It is placed between the focus (f) and the pole.
Therefore f>u>0
For image distance v, we have the mirror formula:
v>0
The image obtained is virtual since it is formed on the right side of the mirror. For u < 0 and v > 0, we can write:
Hence, the formed image is enlarged.
Q 9.16) A small pin fixed on a tabletop is viewed from above from a distance of 50 cm.
By what distance would the pin appear to be raised if it is viewed from the same point
through a 15 cm thick glass slab held parallel to the table? Refractive index of glass =
1.5. Does the answer depend on the location of the slab?
Answer)
The actual depth of the pin, d = 15 cm
Apparent depth of the pin = d’
Refractive index of glass, μ = 1.5
The ratio of actual depth to the apparent depth and the refractive index of the glass are equal.
i.e.
The distance at which the pin appears to be raised =d’-d=15-10=5cm When the angle of incidence is small, the distance is independent of the location of the slab.
Q 9.17)
(a) Figure below shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?
Answer a)
Refractive index of the glass – fibre, μ2 = 1.68
Refractive index of the outer covering, μ2 = 1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i’
The refractive index (μ) of the inner core − outer core interface is given as:
For the critical angle, total internal reflection (TIR) takes place only when i>i’, i.e., i>
59° Maximum angle of reflection,
be the maximum angle of incidence.
The refractive index at the air − glass interface, μ1 =1.68
We know that
Rearranging the equation, we get:
sin imax = 1.68 × sin rmax
Substituting the values in the above equation, we get:
= 1.68 sin 31°
= 1.68 × 0.5150 = 0.8652
Therefore imax = sin−10.8652 = 60°
Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.
Answer b)
If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe, μ1 = Refractive index of air = 1
For the angle of incidence i = 90°
We can write Snell’s law at the air − pipe interface as:
Since i’>r, all incident rays will suffer total internal reflection.
Q 9.18) Answer the following questions:
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver underwater, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?
Answer a) Yes. Real images can be obtained from plane and convex mirrors too. When the light rays converge at a point behind the plane or convex mirror, the object is said to be virtual. The real image of this object is obtained on the screen which is placed in the front of the mirror. This is where the real image is obtained.
Answer b) No To obtain a virtual image, the light rays must diverge. In the eyes, the convex lens help in converging these divergent rays at the retina. This is an example where the virtual image acts as an object for the lens to produce a real image.
Answer c) The diver is in the water while the fisherman is on the land. Air is less dense when compared to the water as a medium. It is mentioned that the diver is viewing the fisherman. This explains that the light rays are travelling from the denser medium to the rarer medium. This means that the refracted rays move away from the normal making fisherman appear taller.
Answer d) Yes; Decrease The reason behind the change in depth of the tank when viewed obliquely is because the light rays bend when they travel from one medium to another. This also means that the apparent depth is less than the near-normal viewing.
Answer e) Yes 2.42 is the refractive index of the diamond while 1.5 is the refractive index of the ordinary glass. Also, the critical angle of the diamond is less than the glass. The sparkling effect of a diamond is possible because of the large cuts in the angle of incidence. Larger cuts ensures that the light entering the diamond is totally reflected from all the faces.
Q 9.19) The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens.
What is the maximum possible focal length of the lens required for the purpose?
Answer)
Distance between the object and the image, d = 3 m
Maximum focal length of the convex lens =fmax For real images, the maximum focal length is given as:
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.
Q 9.20) A screen is placed 90 cm from an object.
The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Answer)
Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
Therefore, the focal length of the convex lens is 21.39 cm.
Q 9.21)
(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement
(a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Answer) Focal length of the convex lens, f1= 30 cm
Focal length of the concave lens, f2 = −20 cm
Distance between the two lenses, d = 8.0 cm
Answer a)
When the parallel beam of light is incident on the convex lens first:
Using lens formula, we can write:
u1 = Object distance
v1 = Image distance 
Therefore v1 = 30cm or a concave lens, the image acts as a virtual object Using lens formula for the concave lens, we can write:
Where,
u2=Object distance
= (30 − d) = 30 − 8 = 22 cm
v2= Image distance
Therefore v2 = −220cm
The parallel incident beam appears to diverge from a point that is 
the centre of the combination of the two lenses.
When the parallel beam of light is incident, from the left, on the concave lens
first:
Using the lens formula, we can write:
u2 = Object distance = −∞
v2 = Image distance
For a convex lens, the image will act as a real object.
Applying lens formula to the convex lens, we have:
u1 = Object distance
= −(20 + d) = −(20 + 8) = −28 cm
v1 = Image distance 
Therefore v2 = −420cm
The parallel incident beam appears to diverge from (420-4) 416cm. The diversion happens from the left of the centre of the combination of the two lenses. The answer is dependent on the side of the combination where the incident beam of light is parallel.
Answer b)
Height of the image, h1 = 1.5 cm
Object distance from the side of the convex lens, u1 = -40 cm
∣u1∣ = 40cm
Using the lens formula:
Therefore v1 = 120cm
Magnification, m= 
Therefore, 3 is the magnification of the convex lens.
The image of the convex lens is an object for the concave lens.
According to the lens formula:
u2 = Object distance
= +(120 − 8) = 112 cm. v2 = Image distance
Therefore v2 
Magnification, m’= 
Hence, the magnification due to the concave lens is 
The magnification produced by the combination of the two lenses is calculated as:
m × m’ = 3 × 20/92 = 60/92 = 0.652
The magnification of the combination is given as:
Hence, the height of the image is 0.98 cm.
Q 9.22) At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face?
The refractive index of the material of the prism is 1.524.
Answer) The incident, refracted and emergent rays associated with a glass prism ABC are shown in the given figure.
Angle of prism, Therefore A = 60° Refractive index of the prism, μ= 1.524
i1 = Incident angle
r1 = Refracted angle
r2 = Angle of incidence at the face AC
e= Emergent angle = 90° According to Snell’s law, for face AC, we can have:
Therefore r1 = A − r2 = 60 − 41 = 19° It is clear from the figure that angle A= r1 + r2
According to Snell’s law, we have the relation:
Q 9.23) A large card divided into squares each of size 1 mm2 is being viewed from a distance of 9 cm through a magnifying glass ( converging lens has a focal length of 9 cm) held close to the eye. Determine:
(a) the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Answer a)
Area of each square, A = 1 mm2
Object distance, u = −9 cm
Focal length of a converging lens, f = 10 cm
The lens formula for the image distance v, can be written as:
Therefore, v = -90 cm
Therefore, area of each square in the virtual image = (10)2 A
= 102 x 1 = 100 mm2
= 1 cm2
Answer b)
The lens has a magnifying power of = 
Answer c)
The magnification in (a) is not the same as the magnifying power in (b).
The magnification magnitude is 
and the magnifying power is 
The two quantities will be equal when the image is formed at the near point (25 cm).
Q 9.24) For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye.
The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
Answer)
Least distance of distinct vision, d = 25 cm
Far point of a normal eye, d’ = ∞
Converging power of the cornea, Pc = 40D Least converging power of the eye-lens, Pe = 20D Eyes use the least converging power to see the objects that are at infinity.
Power of the eye-lens, P = Pc + Pe = 40 + 20 = 60 DPc + Pe = 40 + 20 = 60D
Power of the eye-lens is given as:
To focus an object at the near point, object distance (u) = −d = −25 cm
Focal length of the eye-lens = Distance between the cornea and the retina = Image distance
Hence, image distance, 
Using the lens formula, we can write, 
Where,
f’=focal length
Therefore, Power of the eye-lens = 64 − 40 = 24 D
Hence, 20D to 24D is the range of accommodation of the eye-lens.
Q 9.24)
(a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
Answer a) The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm).
Image distance, v = −d = −25 cm
Focal length, f = 10 cm
Object distance = u
According to the lens formula, we have:
Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.
Answer b)
Magnification 
Answer c)
Magnifying power 
Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.
Q 9.25) What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2?
Would you be able to see the squares distinctly with your eyes very close to the magnifier?
[Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
Answer)
Area of the virtual image of each square, A = 6.25 mm2
Area of each square, A0 = 1 mm2
Hence, the linear magnification of the object can be calculated as:
Focal length of the magnifying glass, f = 10 cm
According to the lens formula, we have the relation:
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.
Q 9.26) Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Answer a)Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.
Answer b) Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.
Answer c) The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.
Answer d) The angular magnification produced by the eyepiece of a compound microscope
fe = Focal length of the eyepiece
It can be inferred that if fe is small, then angular magnification of the eyepiece will be large. The angular magnification of the objective lens of a compound microscope is given as
Answer e) When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much-refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred. The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.
Q 9.27) An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm.
How will you set up the compound microscope?
Answer) Focal length of the objective lens, f0 = 1.25 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation:
The angular magnification of the objective lens (mo) is related to me as: m0me = m
Applying the lens formula for the objective lens:
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
Where,
ve = Image distance for the eyepiece = −d = −25 cm
ue = Object distance for the eyepiece
Separation between the objective lens and the eyepiece = |Ue| + |Vo|
= 4.17 + 7.5
= 11.76cm
Q 9.28) Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
A small telescope has an objective lens of focal length 140cm andan eyepiece of focal length 5.0cm. What is the magnifying power ofthe telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final imageis at infinity)? (b) the final image is formed at the least distance of distinct vision(25cm)?
Answer) The focal length of eyepiece is 5.0 cm and the focal length of objective is 140 cm.
Answer a)
When the telescope is in normal adjustment, its magnifying power is given as,
m= f o f e
Where, f e is the focal length of eyepiece and f o is the focal length of objective.
By substituting the given values in the above expression, we get
m= 140 5 =28
Thus, when the telescope is in normal adjustment, its magnifying power is 28.
Answer b)
When the final image is formed at least distance for distinct vision, the magnifying power of the telescope is given as,
m= f o f e [ 1+ f e d ]
Where, d is the least distance for distinct vision.
By substituting the given values in the above expression, we get
m= 140 5 [ 1+ 5 25 ] =28[ 1+0.2 ] =28×1.2 =33.6
Thus, when the final image is formed at least distance for distinct vision, the magnifying power of the telescope is 33.6.
Q 9.29) Answer the following questions:
(a) For the telescope described in Exercise 9.28 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25cm?
Answer a)
The separation between the objective lens and the eyepiece
= f0 + fe = 140 + 5 = 145 cm.
Answer b)
height of the tower = 100 m; distance = 3 km = 3000 m.
Angle subtended by the tower = 100/3000 = 1/30 rad
Angle subtended by the image produced by the objective
= h/f0 = h/140
∴ h/140 = 1/30
∴ h = 140/30 = 4.7cm
Answer c)
Magnification produced by eyepiece
= (1 + d/fe) = (1 + 25/5) = 6
∴ Height of final image = 14/3 x 6 = 28 cm.
Q 9.30) A Cassegrain telescope uses two mirrors as shown in Fig. 9.30.
Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?
Answer)
Distance between objective mirror and secondary mirror d=20 mm.
Radius of curvature of the large mirror or objective mirror, R 1 =220 mm.
Radius of curvature of the small mirror, R 2 =140 mm
Since, R 1 =220 mm
Thus, focal length of the large mirror
f 1 = R 1 2 = 220 2 =110 mm
Since, R 2 =140 mm
Thus, the focal length of the small mirror is
f 2 = 140 2 =70 mm
The image of an object placed at infinity, formed by the objective mirror will act as a virtual object for the small mirror.
Thus, the object distance for small mirror will be,
u= f 1 −d =110−20 =90 mm
Apply the mirror formula,
1 v = 1 f 2 − 1 u = 1 70 − 1 90 = 1 315
Thus, v=315 mm
Thus, the image will be formed at a distance 315 mmaway from the small mirror.
Q 9.31) Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.36.
A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?
Answer)
Angle of deflection, θ = 3.5°
Distance of the screen from the mirror, D = 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 2θ= 7.0°
The displacement (d) of the reflected spot of light on the screen is given as:
Hence, the displacement of the reflected spot of light is 18.4 cm.
Q 9.32) Figure 9.34 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror.
A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?
Answer)
Focal length of the convex lens, f1 = 30 cm
The liquid acts as a mirror. Focal length of the liquid = f2
Focal length of the system (convex lens + liquid), f = 45 cm
The equivalent focal length of the pair of optical systems that are in contact are given as:
Therefore, f2 = -90 cm
Let μ1 be the refractive index and the radius of curvature of one surface be R. Hence, -R will be the radius of curvature of the other surface.
R can be obtained using the relation : 
Radius of curvature of the liquid on the side of the plane mirror = ∞ Radius of curvature of the liquid on the side of the lens, R = −30 cm
The value of μ2 can be calculated using the relation:
Therefore, 1.33 is the refractive index of the liquid.
That’s it. These were the solutions of NCERT Class 12 Physics Chapter 9 – Ray Optics and Optical Instruments. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.